How do you solve $9 = x + x^{2}$ $9=x+x^2$ ?

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Answer 1 · 2018-06-09T17:21:38

$x = \frac{- 1 \pm \sqrt{37}}{2}$ $x= (-1+-sqrt(37))/2$

$9 = x + x^{2}$ $9=x+x^2$

You can complete the square:

$a x^{2} + b x + c$ $ax^2+bx+c$

a must equal 1

$c = {(\frac{b}{2})}^{2}$ $c=(b/2)^2$

square term $= \frac{b}{2}$ $=b/2$

$x^{2} + x = 9$ $x^2+x = 9$

$a = 1$ $a=1$ so we can proceed.

$c = {(\frac{1}{2})}^{2} = \frac{1}{4}$ $c=(1/2)^2=1/4$

$x^{2} + x + c = 9 + c$ $x^2+x +c = 9+c$

$x^{2} + x + \frac{1}{4} = 9 + \frac{1}{4}$ $x^2+x +1/4 = 9+1/4$

factor the left side:

$(x + \frac{1}{2}) (x + \frac{1}{2}) = 9 + \frac{1}{4}$ $(x+1/2)(x+1/2) = 9+1/4$

${(x + \frac{1}{2})}^{2} = \frac{37}{4}$ $(x+1/2)^2 = 37/4$

now solve:

$\sqrt{{(x + \frac{1}{2})}^{2}} = \pm \sqrt{\frac{37}{4}}$ $sqrt((x+1/2)^2) = +-sqrt(37/4)$

$x + \frac{1}{2} = \pm \sqrt{\frac{37}{4}}$ $x+1/2= +-sqrt(37/4)$

$x = - \frac{1}{2} \pm \frac{\sqrt{37}}{2}$ $x= -1/2+-sqrt(37)/2$

$x = \frac{- 1 \pm \sqrt{37}}{2}$ $x= (-1+-sqrt(37))/2$

graph{x^2+x -9 [-20, 20, -10, 10]}

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